Problem: Let $S = 1 - 2 + 3 - 4 + \cdots + 2009 - 2010$. What is the residue of $S$, modulo 2010?
Solution: Note that $x + 2010 \equiv x \pmod{2010}$. Add 2010 to every negative summand and rearrange the terms to find $S \equiv 0 + 1 + 2 + 3 + \cdots + 2008 + 2009 \pmod{2010}$. The right-hand side is the sum of the integers from 1 to 2010, so $S \equiv \frac{2010 \cdot 2011}{2} \equiv 1005 \cdot 2011 \equiv 1005 \cdot 1 \equiv \boxed{1005} \pmod{2010}$.